R1
t=0
a
b
+
C
V
vc
R
_
Assume the switch in the circuit below has been in terminal (a) for a long time and it is
thrown to terminal (b) at t=0, find the expression for the capacitor voltage vc (t ) for t>0.
After the switch is moved to (b), the circuit is source free. We are interested in the
natural responses of the RC circuit.
The natural response will depend only on the initial energy stored in the capacitor.
We know the capacitor voltage won’t change instantaneously. So we can look at the
circuit before the switch opens.
vc (0  )  vc (0  )  V , initial voltage is equivalent to the source voltage V0  V
The switch goes to b t=0.To find v c (t ) , we use Kirchhoff’s current law (KCL).
Summing the current leaving the node on the top of the circuit:
The circuit is characterized by a first order differential equation.
t>0
b
+
C
vc
R
_
C
dv c v c

0
dt
R
We need to solve the differential equation that describes the circuit.
Rearrange the equation to obtain
We first separate the variable v and t.
dvc
1

dt
vc
RC
Taking the integral of both sides, we have
1
1
vc ( t )
t
vc (0) v di  0  RC dt
c
Consequently we get,
ln vc
vc (t )
1

t
vc (0)
RC

t
Based on the natural log, vc (t )  vc (0)  e t / RC  V0 e  , here   RC
Time constant is an important parameter of the first order circuit.
The smaller the time constant of a circuit, the faster the rate of decay of the response.
Key steps to calculate the step response of a first order circuit:
1. Find the initial voltage V0  vc (0  ) at t  0 
2. Find the time constant   RC
3. vc (t )  V0 e

t
